What is the value of the given limit? 
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How can I prove Infinitesimal Limit 

Let $$\lim_{x\to 0}f(x)=0$$ and $$\lim_{x\to 0}\frac{f(2x)-f(x)}{x}=0$$
Then what is the value of  $$\lim_{x\to 0}\frac{f(x)}{x}$$
 A: For every $\delta > 0$ there exists an $\varepsilon > 0$ such that $|f(2x) - f(x)| < \delta\, |x|$ if $x\neq 0$ and $|x| < \varepsilon$.  Then for all integers $k \geq 0$ $$ \begin{eqnarray}
|f(2x) - f(2^{-k}x)| &=& |f(2x) - f(x) + f(x) - f(2^{-1}x) + \dotsc + f(2^{1-k}x) - f(2^{-k}x)|\\ &<& \delta \,|x| + \delta\, 2^{-1}|x| + \dotsc + \delta \, 2^{-k}|x|\\ &<& 2\delta \, |x| \end{eqnarray}$$ and since $\lim_{k \to \infty}f(2^{-k}x) = 0$ this implies that $|f(2x)| \leq 2\delta \, |x|$ if $|x| < \varepsilon$.  Therefore $$\lim_{x\to 0} \frac{f(x)}{x} = 0.$$
A: If $\displaystyle\lim_{x\to0}\frac{f(x)}{x}=L$ then
$$
\lim_{x\to0}\frac{f(2x)}{x} =  2\lim_{x\to0}\frac{f(2x)}{2x} = 2\lim_{u\to0}\frac{f(u)}{u} = 2L.
$$
Then
$$
\lim_{x\to0}\frac{f(2x)-f(x)}{x} = \lim_{x\to0}\frac{f(2x)}{x} - \lim_{x\to0}\frac{f(x)}{x} =\cdots 
$$
etc.
Later note: What is written above holds in cases in which $\displaystyle\lim_{x\to0}\frac{f(x)}{x}$ exists.  The question remains: If both $\lim_{x\to0}f(x)=0$ and $\lim_{x\to0}(f(2x)-f(x))/x=0$ then does it follow that $\displaystyle\lim_{x\to0}\frac{f(x)}{x}$ exists?
